class Solution:
# time O(2^N) space O(1)
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        res = []
        candidates.sort() # I. this is optional, to optimize speed 
        def dfs(start,target,path):
            if target < 0:  # to optimize, can put this check in for loop and if condition applied, break (assuming sorted)
                return
            if target == 0:
                res.append(path) 
                return
            for i in range(start,len(candidates)):
            #II. optimal solution2:  
                if target - candidates[i] < 0:
                break
                dfs(i,target-candidates[i],path+[candidates[i]])
        dfs(0,target,[])
        return res